Hmm my bad,
Thanks for your replies but I think my example was a little to simple
the actual code I'm using is:
f_haardisolve<-function(v_dataset){
#pad data to make length a power of 2
v_dataset<-f_paddata(v_dataset)
l_cooef<-list() #holder for cooefficents
i_count<-1 #identity counter
while(length(v_dataset)>0){
#seperate odd and even points
v_dataset<-f_splitpoints(v_dataset)
v_dataset<-f_haarpredict(v_dataset)
v_dataset<-f_haaruplift(v_dataset)
str_idx<-paste('c',i_count,sep='')
l_cooef
$str_idx<-v_dataset[c((length(v_dataset)/2)+1:(length(v_dataset)/2))]
#should be no wrap on the previous line
v_dataset<-v_dataset[c(1:length(v_dataset)/2)]
i_count<-i_count+1
}
l_cooef
}
In this particular case I could probably calculate the number of
iterations and use l_cooef<-names() but more is there a more generic
method for adding another column of data?
On Tue, 2005-27-09 at 10:39 -0400, tom wright wrote:
> Can someone please show me what I need to get something like this to
> work
>
> for(a in c(1:5)){
> data$a<-c(a:10)
> }
>
> so that i end up with a structure
> data$1<-[1,2,3,4,5,6,7,8,9,10]
> data$2<-[2,3,4,5,67,8,9,10]
> data$3<-[3,4,5,67,8,9,10]
> data$4<-[4,5,67,8,9,10]
> data$5<-[5,67,8,9,10]
>
> thanks loads
> Tom
>
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